Replace x x and substitute

Algebra Level 3

x 2 + x = 1 , x 6 + 2 x 5 + 2 x 4 + 2 x 3 + 2 x 2 + 2 x + 1 = ? \large x^2+x=1 \quad, \quad{x^6+2x^5+2x^4+2x^3+2x^2+2x+1} = \ ?


The answer is 4.

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Adam Phúc Nguyễn by Adam Phúc Nguyễn , 20, Vietnam

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2 solutions

x 6 + 2 x 5 + 2 x 4 + 2 x 3 + 2 x 2 + 2 x + 1 {x^6+2x^5+2x^4+2x^3+2x^2+2x+1}

= x 2 ( x 4 + 2 x 3 + x 2 ) + ( x 4 + 2 x 3 + x 2 ) x 2 + 2 x + 1 =x^2(x^4+2x^3+x^2)+(x^4+2x^3+x^2)x^2+2x+1

= x 2 ( x 2 + x ) 2 + ( x 2 + x ) 2 + x + 2 {=x^2(x^2+x)^2+(x^2+x)^2+x+2}

= x 2 + x + 3 = 1 + 3 = 4 {=x^2+x+3=1+3=\boxed{4}}

14 Helpful 0 Interesting 0 Brilliant 0 Confused
Ben Habeahan
Aug 20, 2015

Let x 6 + 2 x 5 + 2 x 4 + 2 x 3 + 2 x 2 + 2 x + 1 = p x^6+2x^5+2x^4+2x^3+2x^2+2x+1=p so we have, x 6 + 2 x 5 + 2 x 4 + 2 x 3 + 2 x 2 + 2 x + 1 − p = 0 … ( ∗ ) x^6+2x^5+2x^4+2x^3+2x^2+2x+1-p=0 \dots(*) And from x 2 + x = 1 , x^2+x=1, we have x 2 + x − 1 = 0 … ( ∗ ∗ ) . x^2+x-1=0 \dots(**). \\ Divide ( ∗ ) (*) by ( ∗ ∗ ) (**) using long division and remainder is 4 − p 4-p \\ So from theorema factor, remainder must be 0 0 4 − p = 0 p = 4 \\ 4-p=0 \\ p= \boxed{4}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

" and remain is 4 − p 4-p So the remain must be 0" I don't actually understand this part.

Adam Phúc Nguyễn - 5 years, 9 months ago

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4 − p 4-p means remainder from, ( ∗ ) ( ∗ ∗ ) = x 4 + x 3 + 2 x 2 + x + 3 + 4 − p ( ∗ ∗ ) \frac{(*)}{(**)}=x^4+x^3+2x^2+x+3+ \frac{4-p}{(**)} \\

Ben Habeahan - 5 years, 9 months ago

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Oh, thanks.

Adam Phúc Nguyễn - 5 years, 9 months ago

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